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The magnitude of the acceleration of either block is 1.4 m/s²
The diagram referred to in the question is shown below:
The coefficient of kinetic friction between the surface and the larger block is:
[tex]\mu_1 = 0.25[/tex]
The coefficient of kinetic friction between the surface and the smaller block is:
[tex]\mu_2 = 0.40[/tex]
The mass of the larger block, [tex]m_1 = 3M[/tex]
M = 1.0 kg
[tex]m_1 = 3(1)\\m_1 = 3 kg[/tex]
The mass of the smaller block, [tex]m_2 = 2M[/tex]
[tex]m_2 = 2(1)\\m_2 = 2 kg[/tex]
The force, F = 22 N
Calculate the acceleration using the equation below:
[tex]F = (\mu_1m_1+\mu_2m_2)g + (m_1+m_2)a\\22 = 9.81[(0.25 \times 3) + (0.4 \times 2)] + (3 + 2)a\\22 = 9.81(0.75+0.8) + 5a\\22 = 9.81(1.55) + 5a\\22 = 15.2055 + 5a\\5a = 22 - 15.2055\\5a = 6.7945\\a = \frac{6.7945}{5} \\a = 1.3589\\a = 1.4 m/s^2[/tex]
The magnitude of the acceleration of either block is 1.4 m/s²
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